# Sequence of measurable subsets of [0,1] (Lebesgue measure, Measurable)

1. May 29, 2013

### ChemEng1

1. The problem statement, all variables and given/known data
Let $\left\{E_{k}\right\}_{k\in N}$ be a sequence of measurable subsets of [0,1] satisfying $m\left(E_{k}\right)=1$. Then $m\left(\bigcap^{\infty}_{k=1}E_{k}\right)=1$.

2. Relevant equations
m denotes the Lebesgue measure.
"Measurable" is short for Lebesgue-measurable.

3. The attempt at a solution
My intuition was to construct 2 disjoint and uncountable subsets of an uncountable set that all have measure equal to 1.

After much struggling, I don't think this is possible. To be able to construct such a counterexample would mean there is a mapping of each subset to the natural numbers which would make them both countable.

Are there non-constructive approaches to building this counterexample? My attempts haven't yielded anything meaningful. Each example (rational and irrational, transcendental and algebraic) I've considered ends up with measure 1 and measure 0 subsets instead of both measure 1.

I am not sure if my counterexample building skills are lacking or I just need to adjust my intuition. Any help would be appreciated.

2. May 29, 2013

### Ray Vickson

What is the measure of the complement (in [0,1]) of $\cap_{k=1}^{\infty} E_k$?

3. May 29, 2013

### ChemEng1

If I can construct a counterexample of 2 disjoint and noncountable subsets on [0,1] of measure 1, then the measure of the complement of intersections would be 1. However, I have not been able to find such a counterexample. The answer based on what I have considered is 0.

4. May 29, 2013

### Ray Vickson

So, you do not intend to answer my question. Fair enough, but that's it for me.

5. May 29, 2013

### ChemEng1

Your question is closely related to the question I am posing. If I could simply answer it, then I wouldn't've started the thread.

6. May 29, 2013

### D H

Staff Emeritus
Try again, but this time answer the question. Start by answering this question: What is the complement (in [0,1]) of $\cap_{k=1}^{\infty} E_k$? When you have answered this, can you find an upper bound on the measure of that complement?

7. May 29, 2013

### ChemEng1

By DeMorgan, $C\left(\cap_{k=1}^{\infty} E_k\right)=\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}$

A lower bound immediately comes to mind. I'm still stewing on how to get an upper bound to pop out.

Lower Bound:
Consider: $m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]$. $m\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]= m\left[0,1\right]-m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)\geq m\left[0,1\right]-\Sigma_{k=1}^{\infty} m\left(E^{c}_{k}\right)=1-0=1$

8. May 29, 2013

### Office_Shredder

Staff Emeritus
You have a set which is a subset of [0,1] and has measure bigger than or equal to 1, and you're not sure what its measure is?

9. May 29, 2013

### Ray Vickson

In Step 1 you did not follow through all the way: if $E = \cap_k E_k,$ then
$$E^c = \bigcup_k E_k^c$$

10. May 30, 2013

### ChemEng1

Yep. I messed that up too. One more time.

By DeMorgan, $C\left(\cap_{k=1}^{\infty} E_k\right)=C\left[\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)^{c}\right]=\bigcup_{k=1}^{\infty}E^{c}_{k}$

Consider: $m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)$. $m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)\leq \Sigma_{k=1}^{\infty} m\left(E^{c}_{k}\right)=0$

Therefore, $m\left(\bigcup_{k=1}^{\infty}E^{c}_{k}\right)=0\Rightarrow m\left(\cap_{k=1}^{\infty}E_{k}\right)=1$

11. May 30, 2013

### Ray Vickson

Easier: from $E = \cap_k E_k$ we have $E^c = \cup_k E_k^c$ (no need for all those intermediate steps). Then $$0 \leq m(E^c) \leq \sum_k m(E_k^c) = \sum_k 0 =0,$$ where the $`\leq '$ is from some standard property or result, depending on exactly what approach was used in the course/notes/textbook.

Last edited: May 30, 2013
12. May 30, 2013

### ChemEng1

I like the sandwhich approach better than mine.

Thanks for the pointers. I really lost the forest from the trees on this problem.

I got it stuck in my head that perhaps the irrationals were "dense enough" to create 2 subsets with the same measure. It didn't dawn on me to use L.measure properties to disprove the existence of that set.